Question

Question 98(iv)

Simplify
$(2^5÷2^8)\times 2^{-7}$

Answer 1

Using the laws of exponent,
$a^m÷a^n=(a{)}^{m-n}and{a}^m\times a^n=(a{)}^{m+n}$

$\therefore (2^5÷2^8)\times (2{)}^{-7}=(2{)}^{5-8}\times (2{)}^{-7}$


$=(2{)}^{-3}\times (2{)}^{-7}$

$=(2{)}^{-3+(-7)}$

$=(2{)}^{-3-7}$

$=(2{)}^{-10}=\frac{1}{2^{10}}=\frac{1}{1024}[\because a^{-m}=\frac{1}{a^m}]$